We want to find all solutions in integers to the equation x2+2=y3. A quick bit of thought gives x=±5, y=3, but are there any others? The answer to this is in fact "no". Let's see why.

First, Z(i√2) is a Euclidean domain with the Euclidean valuation f(a+ib√2)=a2+2b2 (this is just the square norm of the complex number a+ib√2, so this
is easy to check). Hence it's a UFD. Note that the only units are ±1,
so I'll ignore signs where relevant from now on. We can factor our
equation as y3=(x+i√2)(x-i√2). If c divides both of the
factors on the RHS, it divides their difference 2i√2, so it must be
one of 1, 2, i√2, or 2i√2. Now it's easy to see that x must be odd
(just reduce the original equation mod 4). Thus c must be 1. That is,
the two factors are coprime, and since their product is a cube
(and the units are both cubes), each individual factor must also be a
cube. Putting x+i√2=(g+ih√2)3, we obtain 1=3g2h-2h3, so h=±1. If h=-1, there's no integer g which works, so h=1 and g=±1. Thus x=g3-6gh2=±5, and thus y=3. Both these solutions work, and they are the only ones.